I am trying to stub the following code
async function logUpdate(client) {
const results = await client.query(query.toParam());
const { count, timestamp } = results.rows[0];
await db.updateDatasourceLogs(destdb, DB.src, TABLES.src, timestamp, count);
}
This is the following code i am using to stub the above code
fakeClient = {
query: sinon.stub().resolves(fakeRows),
};
const rowData = {
count: 1,
timestamp: ''
};
fakeRows = {
rows: sinon.stub().returns([rowData]),
};
fakeSequel = {
useFlavour: sinon.stub().returns(toParam: () => false,),
};
I am getting an error for destructuring
TypeError: Cannot destructure property count of 'undefined' or 'null'.
at line
const { count, timestamp } = results.rows[0];
how to stub the above line?
If we look at the body of your logUpdate function, we see it begins with the following two lines:
const results = await client.query(query.toParam());
const { count, timestamp } = results.rows[0];
This code says:
client.query and assign
it to a variable called results.results is an Object with a property called rows which is an
Array whose 0th element should be an Object with count and
timestamp properties - which we destructure into local variables.This implies that the value of results looks like:
{
"rows": [
{
"count": 1
"timestamp": "0"
}
]
}
However, in our stubbing code, we have the following:
fakeRows = {
rows: sinon.stub().returns([rowData]),
};
Which says that fakeRows is an Object with a rows property whose value is a function that returns [rowData].
If we were to implement this Object without sinon, it would look like:
{
"rows": function () {
return [
{
"count": 1
"timestamp": "0"
}
];
}
}
Notice the difference between the structure that logUpdate is expecting and what fakeRows actually provides. Specifically, logUpdate expects results to have a rows Array, but, with our stubbed code, it has a rows Function!
We can fix this simply by having fakeRows reference our rowData directly:
const fakeRows = {
rows: [rowData]
};